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Prove that Tan a + Tan B Tan a − Tan B = Sin ( a + B ) Sin ( a − B ) - Mathematics

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प्रश्न

Prove that
\[\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin \left( A + B \right)}{\sin \left( A - B \right)}\]

संक्षेप में उत्तर
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उत्तर

\[LHS = \frac{\tan A + \tan B}{\tan A - \tan B}\]
\[ = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}\]
\[ = \frac{\frac{\sin A \cos B + \cos A\sin B}{\cos A \cos B}}{\frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B}}\]
\[ = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}\]
\[ = \frac{\sin\left( A + B \right)}{\sin\left( A - B \right)} \]
 = RHS
Hence proved .

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ १९]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 10 | पृष्ठ १९

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